Asked by Steve
In a physics lab, a small cude slides down a frictionless incline, and elastically strikes a cube at the bottom of the inclince that is only one-half its mass. If the incline is 30 cm high and the table is 90 cm off the floor, where does each cube land?
I got the V0 to be 2.43 from mhg = 1/2mv^2 but after setting up M1V1=M1V1' + M2V2' i don't know what to do
I got the V0 to be 2.43 from mhg = 1/2mv^2 but after setting up M1V1=M1V1' + M2V2' i don't know what to do
Answers
Answered by
drwls
The speed of the sliding cube at impact is V = sqrt(2gH) = 2.42 m/s
Next you have to figure out the velocities of the sliding cube (mass M) and the impacted cube (Mass M/2), after impact. You need to write equations of both momentum and kinetic energy conservation. You will find, if you do it right, that the smaller cube travels at 4V/3 and the larger cube at V/3.
Finally, use those horizontal velocities and the height of the table to predict where they land.
Next you have to figure out the velocities of the sliding cube (mass M) and the impacted cube (Mass M/2), after impact. You need to write equations of both momentum and kinetic energy conservation. You will find, if you do it right, that the smaller cube travels at 4V/3 and the larger cube at V/3.
Finally, use those horizontal velocities and the height of the table to predict where they land.
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